This problem requires you to use both kinds of special right triangles.
A 30°-60°-90° triangle has three angles with measures of 30°, 60°, and 90°. The side lengths are related by a special property, such that the shortest leg (always across from the 30° angle) is y units, the hypotenuse (always across from the 90° angle in every right triangle) is 2y units, and the third leg is [tex]y \sqrt{3} [/tex] units. (I use y because this is a ratio depending on what we have for y. We can write the simplest ratio as 1:2:√3.)
So it looks like triangle RST is one of these triangles.
If [tex]2 \sqrt{3} [/tex] is the leg across from the 60°-angle, then we have 2√3 = y√3, so y = 2. That means our hypotenuse is 2y = 4.
Now the hypotenuse is one of the legs of the larger right triangle, QRT.
This is a right isosceles triangle, or a 45°-45°-90° triangle. This is given such that the two congruent legs are each y units long, and the hypotenuse is [tex]y \sqrt{2} [/tex] (or simplified, 1:1:√2).
But we just care about the length of x (side RQ), which we see is the leg that is congruent to RT. The length of RT is 4, so the length of RQ is also 4. Therefore, x = 4.
